## skyhimself75 3 years ago what is the solution of 5|x+1|+6=21

1. amistre64

how would we start it?

2. Alexis1994_13

5|χ+1|=15 => |x+1| = 3.

3. amistre64

im sure as long as you just get an answer; you will continue to type in the same kind of problem over and over again :)

4. Llort

assuming x is real: $5 \sqrt((x+1)^2)+6 = 21$ otherwise: |x+1|=3

5. rauch

2

6. Llort

don't just give it to him.

7. skyhimself75

i don't know how to start it absolute value i just don't understand it

8. amistre64

the absolute value is the last thing to work on; how do you get that absolute value all by itself?

9. rauch

5(x+1)+6=21 (5x+5)+6=21 5x+6=16 5x=10 x=2

10. Llort

looking at this question, what is the other solution for x if it was not an absolute value?

11. skyhimself75

i'm seriously terrible in math i really don't know:[

12. amistre64

the term "absolute value" means that there is a quantifiable value; something that can be measured. length, height, area, volume, distance are all examples of abslute values

13. jgeorge123

dont jus give him the answer he is probably usin this site for a summer school!!!

14. amistre64

in other words; an absolute value doesnt care about what direction the sign takes us in; it only cares about a quantifiable measure

15. Llort

|a| is always positive, you discount the negative answer.

16. polpak

sky, you isolate the absolute value just like you would isolate a variable. Llort, no you do not 'discount' the negative answer. It is a valid solution.

17. amistre64

|-12| has the same absolute value as |12|

18. amistre64

as a result; if we are determining any value for inside of the absolute value notation; we have to account for the positive and the negative results that occur within it

19. polpak

So if you have $5|x+1| + 6 = 21$ You want to isolate the term with the absolute value in it first. You do this by subtracting 6 from both sides of the equation.

20. polpak

What do you get when you subtract the 6?

21. skyhimself75

-2?

22. polpak

No, $$5|x+1| + 6 - 6 = 21 -6$$ $\implies 5|x+1| + 0 = 15$$\implies 5|x+1| = 15$

23. polpak

So now lets say that |x+1| is something, lets call it k.. So we have $$5k = 15$$$\implies \frac{1}{5}\cdot5k = \frac{1}{5}15$$\implies k = 3$ And since we said k was |x+1| we have $|x+1| = 3$So what values can x have and still make this equation true?

24. Llort

I am assuming the other answer is x=-4

25. polpak

x =-4 is one solution yes, because |-4 + 1 | = |-3| = 3